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hdu 2602 Bone Collector

科牛 2022-09-07 阅读 231

Bone Collector


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 61757    Accepted Submission(s): 25729


Problem Description


Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …

The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?



hdu 2602 Bone Collector_java



 



Input


The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.


 



Output


One integer per line representing the maximum of the total value (this number will be less than 2 31).


 



Sample Input


1 5 10 1 2 3 4 5 5 4 3 2 1


 



Sample Output


14

看懂他在表达什么意思了,然后自己推一遍,然后这个题迎刃而解。下面是我的代码!

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1005];
int b[1005];
int dp[1005][1005];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
int i,n,v,j;
scanf("%d%d",&n,&v);
for(i=1; i<=n; i++)
scanf("%d",&a[i]);
for(i=1; i<=n; i++)
scanf("%d",&b[i]);
for(i=0; i<=v; i++)
{
if(i<b[n])
dp[n][i]=0;
else
dp[n][i]=a[n];
}
for(i=n-1; i>=1; i--)
{
for(j=0; j<=v; j++)
{
if(j<b[i])
dp[i][j]=dp[i+1][j];
else
dp[i][j]=max(dp[i+1][j],dp[i+1][j-b[i]]+a[i]);
}
}
printf("%d\n",dp[1][v]);
/* for(i=0;i<=n;i++)
{
for(j=0;j<=v;j++)
printf("%d ",dp[i][j]);
printf("\n");
}*/
}
return 0;
}



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